# A charge of 40 C passes through a circuit every 8 s. If the circuit can generate 42 W of power, what is the circuit's resistance?

Feb 10, 2016

$R = 1.68 \Omega$

#### Explanation:

The current, $I$, is the charge passing through per unit time ($\frac{Q}{C}$).

So,

I = frac{40"C"}{8"s"} = 5"A"

The power arises from the fact that charges are moving over a potential difference.

$P = I V$

From Ohm's Law, we know that $V = I R$ for an ideal resistance.

Therefore,

$P = {I}^{2} R$

42"W" = (5"A")^2R

$R = 1.68 \Omega$

Feb 10, 2016

$1.68 \Omega$

#### Explanation:

We'll need 2 formulae to answer this. namely:

$Q = I t$ and $P = {I}^{2} R$

Where Q is charge (40C), t is time (8s), P is power (42W), I is current and R is resistance.

Begin by using the first formula to obtain the current:

$I = \frac{Q}{t} = \frac{40}{8} = 5 A$

Now use the second formula to get the resistance:

$R = \frac{P}{I} ^ 2 = \frac{42}{5} ^ 2 = \frac{42}{25} = 1.68 \Omega$